Lue Vang
Mr. Kellogg
AP Physics - Pd. 7
20 November 2011
Performed On: November 2011
Lab Partners: Arielle Roth
Projectile Motion Lab Purpose:
Not Required
Background:
Not Required
Hypothesis:
Not Required
Apparatus:
Not Required
Procedures:
Not Required
Data:
The curve of the trajectory of the ball is a negatively parabolic curve because the y-component starts with a positive velocity, but gradually decreasing due to an acceleration of -9.8m/s.
Also, the distance between each point's x-component is approximately equal because because (x) has a acceleration of 0.
In other words, the x-component of velocity is the constant, thus the change in (x) is the same throughout the whole experiment.
The graph for position(x) is linear. This suggests that the velocity of (x) is constant, not experiencing any acceleration.
Like the trajectory graph, the position(y) vs. time graph is also negatively parabolic. This hints at the fact that the trajectory graph's parabolic quality originates from the parabolic quality of its y-component. Also, it's parabolic nature shows a negative change in velocity, a negative acceleration.
The velocity(x) vs. time graph strangely shows a negatively linear model. The graph should be a horizontal line, for there should've been zero change in velocity, zero accleration. The strange occurance may be due to human error, for the person plotting the data may not have been accurately tracing the object's projectile path.
The graph of velocity(y) vs. time shows a negatively linear curve, thus there is a negative acceleration . The graph shows that the object is slowing down in one direction, then changes direction when the plot crosses the x-axis and begins speeding up.
Analysis:
Using the Pythagorean Theorem, it is possible to use the (x) and (y) components of the velocity graph to calculate the velocity of the ball at a certain point in time.
Interestingly, the resultant velocity of the two sample times are relatively equal around 3 m/s. Nonetheless, the solution is mere coincidence because velocity does not remain constant throughout the path of a projectile object. For example, at 0.8seconds, the velocity is about 4.233 m/s
Mr. Kellogg
AP Physics - Pd. 7
20 November 2011
Performed On: November 2011
Lab Partners: Arielle Roth
Projectile Motion Lab
Purpose:
Not Required
Background:
Not Required
Hypothesis:
Not Required
Apparatus:
Not Required
Procedures:
Not Required
Data:
The curve of the trajectory of the ball is a negatively parabolic curve because the y-component starts with a positive velocity, but gradually decreasing due to an acceleration of -9.8m/s.
Also, the distance between each point's x-component is approximately equal because because (x) has a acceleration of 0.
In other words, the x-component of velocity is the constant, thus the change in (x) is the same throughout the whole experiment.
The graph for position(x) is linear. This suggests that the velocity of (x) is constant, not experiencing any acceleration.
Like the trajectory graph, the position(y) vs. time graph is also negatively parabolic. This hints at the fact that the trajectory graph's parabolic quality originates from the parabolic quality of its y-component. Also, it's parabolic nature shows a negative change in velocity, a negative acceleration.
The velocity(x) vs. time graph strangely shows a negatively linear model. The graph should be a horizontal line, for there should've been zero change in velocity, zero accleration. The strange occurance may be due to human error, for the person plotting the data may not have been accurately tracing the object's projectile path.
The graph of velocity(y) vs. time shows a negatively linear curve, thus there is a negative acceleration . The graph shows that the object is slowing down in one direction, then changes direction when the plot crosses the x-axis and begins speeding up.
Analysis:
Using the Pythagorean Theorem, it is possible to use the (x) and (y) components of the velocity graph to calculate the velocity of the ball at a certain point in time.
Pythagorean Theorem:
a^2 + b^c = c^2or
Vx^2 + Vy^c = VR^2
Examples:
At t = 0.2 sec (way up)
Vx = 2.6741 ... ... ... Vy = 1.524
VR = sqrt(Vx^2 + Vy^2)
VR = sqrt(2.6741^2 + 1.534^2)
VR = 3.083 m/s
At t = 0.6 sec (way down)
Vx = 2.3923... ... ... Vy = -1.882
VR = sqrt(Vx^2 + Vy^2)
VR = sqrt(2.3923^2 + (-1.882)^2)
VR = 2.966 m/s
Interestingly, the resultant velocity of the two sample times are relatively equal around 3 m/s. Nonetheless, the solution is mere coincidence because velocity does not remain constant throughout the path of a projectile object. For example, at 0.8seconds, the velocity is about 4.233 m/s